In dragons, red color (R) is dominant to white (r), blue flame (B) is dominant to orange flame (b), and small fangs (F) are dominant to large fangs (f)...Imagine a dragon, RrFfBb is test crossed and has 1000 offspring with the following phenotypes...

Red with orange flame and small fangs =9Red with orange flame and large fangs=43Red with blue flame and small fangs =140Red with blue flame ad large fangs =305White with blue flame and small fangs=42white with blue flame and large fangs=6white with orange flame and small fangs=310white with orange flame and large fangs=145

Which genes are linked? we need to find recombination frequency for each of the combos of pairs of genes.

One solution is to look at pair of characters instead of triplets. So you can estimate the linkage frequency between each pair of gene. It is relatively easy and quick for a pair to compare the observed frequency and compare it to th expected frequency in case of independance.

Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Test cross means crossing subject1 with dominant phenotype and unknown genotype (even though you do know it in this case) to subject2 with recessive phenotype (therefore, known genotype – homozygous recessive). The objective is to prove that subject1 is homozygous or heterozygous for the trait.

So you are cross is:RrBbFf x rrbbff

Now, all of the offspring will receive the same (rbf) recessive set of genes from one parent. So you can convert all resulting phenotypes to the sets of genes inherited from RrBbFf.

Red with orange flame and small fangs RbF =9Red with orange flame and large fangs Rbf =43Red with blue flame and small fangs RBF =140Red with blue flame ad large fangs RBf =305White with blue flame and small fangs rBF =42white with blue flame and large fangs rBf =6white with orange flame and small fangs rbF =310white with orange flame and large fangs rbf =145

Now, the most prevalent genotypes are those of the parental combination (no recombination). As you can see your RrBbFf parent produced RBf and rbF gametes most commonly. The second thing you can look at is the general picture. If none of the genes are linked, all phenotypes should occur with the same frequency. As you can see in this case each set of gametes with the same frequency constitute pairs forming parental genotype (RBF + rbf = RrBbFf). Thus, you can deduce that all 3 genes are linked.

Since now you are looking at the 3 gene linkage, you need to establish order of genes in sequence. You already established that RBf and rbF are non-recombinant, now you need to look at the opposite, a double recombinant. Those should be the rarest of all (it requires 2 recombination effects close by to each other). As you can see, RbF and rBf fit this description.

Now, if RBf and rbF are non-recombinant and RbF and rBf are double recombinant, you can establish which of the genes is in the middle. Try different combinations. You will find that B-R-f (or f-R-B in reverse) and b-r-F are the only solutions.

The only thing left, is to do the math and calculate distances B-R, R-f, and B-f and see if answers support those conclusions.